Using differentials and given $h=fcircvec g$ find $h(1.02,1.99)$
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Using differentials find approximately $h(1.02,1.99)$ using that
$$h=fcircvec g,quad f(u,v)=3u+v^2,quadvec g(1,2)=(3,6),quad D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright).$$
What does it mean by "Using differentials"?
Anyway let $h(x,y)$ a differentiable function, because both $f$ and $vec g$ are differentiable because the first one is a composition of sums and cuadratic, and the second because that has given us the jacobian, therefore their composition is also differentiable $colorred(textin only one point or in the domain of h)$?
So we must find $$h(x,y)=h(x_0,y_0)+h'_x(x_0,y_0)(x-x_0)+h'_y(x_0,y_0)(y-y_0).$$
Using the composition of functions I get
$$begincasesh'_x&=f'_ucdot g'_x+f'_vcdot g'_x\h'_y&=f'_ucdot g'_y+f'_vcdot g'_y,endcases$$ but I am not able to continue because I don't know how to evaluate the derivatives of $vec h$ (for $f$ yes, finding its gradient). Also I do not know what is the value of $h(x_0,y_0)$.
Can anyone help me, please?
Thank you!
multivariable-calculus partial-derivative
asked Aug 7 at 19:09
manooooh
390214
 |Â
show 4 more comments
up vote
1
down vote
favorite
Using differentials find approximately $h(1.02,1.99)$ using that
$$h=fcircvec g,quad f(u,v)=3u+v^2,quadvec g(1,2)=(3,6),quad D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright).$$
What does it mean by "Using differentials"?
Anyway let $h(x,y)$ a differentiable function, because both $f$ and $vec g$ are differentiable because the first one is a composition of sums and cuadratic, and the second because that has given us the jacobian, therefore their composition is also differentiable $colorred(textin only one point or in the domain of h)$?
So we must find $$h(x,y)=h(x_0,y_0)+h'_x(x_0,y_0)(x-x_0)+h'_y(x_0,y_0)(y-y_0).$$
Using the composition of functions I get
$$begincasesh'_x&=f'_ucdot g'_x+f'_vcdot g'_x\h'_y&=f'_ucdot g'_y+f'_vcdot g'_y,endcases$$ but I am not able to continue because I don't know how to evaluate the derivatives of $vec h$ (for $f$ yes, finding its gradient). Also I do not know what is the value of $h(x_0,y_0)$.
Can anyone help me, please?
Thank you!
multivariable-calculus partial-derivative
asked Aug 7 at 19:09
manooooh
390214
1
You provide $quad D_vec g$. But in which $(x,y)$ value is it?
– mathcounterexamples.net
Aug 7 at 19:18
1
Using differentials means: using the approximation $f(x+delta)=f(x)+delta f'(x)$
– N74
Aug 7 at 19:20
@N74 thank you. Hmmm... and in the case for $2$ variables? Something like $h(x+delta_1,y+delta_2)$?
– manooooh
Aug 7 at 19:27
1
You have already the right expression for $h$
– N74
Aug 7 at 19:33
1
The vector form of my first comment... the derivative is replaced by the gradient in vector spaces
– N74
Aug 7 at 20:03
 |Â
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Using differentials find approximately $h(1.02,1.99)$ using that
$$h=fcircvec g,quad f(u,v)=3u+v^2,quadvec g(1,2)=(3,6),quad D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright).$$
What does it mean by "Using differentials"?
Anyway let $h(x,y)$ a differentiable function, because both $f$ and $vec g$ are differentiable because the first one is a composition of sums and cuadratic, and the second because that has given us the jacobian, therefore their composition is also differentiable $colorred(textin only one point or in the domain of h)$?
So we must find $$h(x,y)=h(x_0,y_0)+h'_x(x_0,y_0)(x-x_0)+h'_y(x_0,y_0)(y-y_0).$$
Using the composition of functions I get
$$begincasesh'_x&=f'_ucdot g'_x+f'_vcdot g'_x\h'_y&=f'_ucdot g'_y+f'_vcdot g'_y,endcases$$ but I am not able to continue because I don't know how to evaluate the derivatives of $vec h$ (for $f$ yes, finding its gradient). Also I do not know what is the value of $h(x_0,y_0)$.
Can anyone help me, please?
Thank you!
multivariable-calculus partial-derivative
asked Aug 7 at 19:09
manooooh
390214
Using differentials find approximately $h(1.02,1.99)$ using that
$$h=fcircvec g,quad f(u,v)=3u+v^2,quadvec g(1,2)=(3,6),quad D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright).$$
What does it mean by "Using differentials"?
Anyway let $h(x,y)$ a differentiable function, because both $f$ and $vec g$ are differentiable because the first one is a composition of sums and cuadratic, and the second because that has given us the jacobian, therefore their composition is also differentiable $colorred(textin only one point or in the domain of h)$?
So we must find $$h(x,y)=h(x_0,y_0)+h'_x(x_0,y_0)(x-x_0)+h'_y(x_0,y_0)(y-y_0).$$
Using the composition of functions I get
$$begincasesh'_x&=f'_ucdot g'_x+f'_vcdot g'_x\h'_y&=f'_ucdot g'_y+f'_vcdot g'_y,endcases$$ but I am not able to continue because I don't know how to evaluate the derivatives of $vec h$ (for $f$ yes, finding its gradient). Also I do not know what is the value of $h(x_0,y_0)$.
Can anyone help me, please?
Thank you!
multivariable-calculus partial-derivative
asked Aug 7 at 19:09
manooooh
390214
edited Aug 7 at 19:41
asked Aug 7 at 19:09
manooooh
390214
asked Aug 7 at 19:09
manooooh
390214
asked Aug 7 at 19:09
manooooh
390214
390214
1
You provide $quad D_vec g$. But in which $(x,y)$ value is it?
– mathcounterexamples.net
Aug 7 at 19:18
1
Using differentials means: using the approximation $f(x+delta)=f(x)+delta f'(x)$
– N74
Aug 7 at 19:20
@N74 thank you. Hmmm... and in the case for $2$ variables? Something like $h(x+delta_1,y+delta_2)$?
– manooooh
Aug 7 at 19:27
1
You have already the right expression for $h$
– N74
Aug 7 at 19:33
1
The vector form of my first comment... the derivative is replaced by the gradient in vector spaces
– N74
Aug 7 at 20:03
 |Â
show 4 more comments
1
You provide $quad D_vec g$. But in which $(x,y)$ value is it?
– mathcounterexamples.net
Aug 7 at 19:18
1
Using differentials means: using the approximation $f(x+delta)=f(x)+delta f'(x)$
– N74
Aug 7 at 19:20
@N74 thank you. Hmmm... and in the case for $2$ variables? Something like $h(x+delta_1,y+delta_2)$?
– manooooh
Aug 7 at 19:27
1
You have already the right expression for $h$
– N74
Aug 7 at 19:33
1
The vector form of my first comment... the derivative is replaced by the gradient in vector spaces
– N74
Aug 7 at 20:03
1
1
You provide $quad D_vec g$. But in which $(x,y)$ value is it?
– mathcounterexamples.net
Aug 7 at 19:18
You provide $quad D_vec g$. But in which $(x,y)$ value is it?
– mathcounterexamples.net
Aug 7 at 19:18
1
1
Using differentials means: using the approximation $f(x+delta)=f(x)+delta f'(x)$
– N74
Aug 7 at 19:20
Using differentials means: using the approximation $f(x+delta)=f(x)+delta f'(x)$
– N74
Aug 7 at 19:20
@N74 thank you. Hmmm... and in the case for $2$ variables? Something like $h(x+delta_1,y+delta_2)$?
– manooooh
Aug 7 at 19:27
@N74 thank you. Hmmm... and in the case for $2$ variables? Something like $h(x+delta_1,y+delta_2)$?
– manooooh
Aug 7 at 19:27
1
1
You have already the right expression for $h$
– N74
Aug 7 at 19:33
You have already the right expression for $h$
– N74
Aug 7 at 19:33
1
1
The vector form of my first comment... the derivative is replaced by the gradient in vector spaces
– N74
Aug 7 at 20:03
The vector form of my first comment... the derivative is replaced by the gradient in vector spaces
– N74
Aug 7 at 20:03
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Supposing that
$$D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright)$$
you also have $D_f(3,6)= (3,12)$.
Now you can write $(1.02,1.99)=(1+0.02,2.0-0.01)=(1,2)+(0.02,-0.01)=(1,2)+(h,k)$.
Based on that, you can use the chain rule (which you used in your question using coordinates), namely
$$D_h(1,2)=D_f(3,6) circ D_vec g(1,2)$$
To get
$$h(1.02,1.99)approx h(1,2)+D_h(1,2).(h,k)=45+(3,12)left[beginpmatrix2 &1\
3&5endpmatrixbeginpmatrix0.02\
-0.01endpmatrixright]=45.21$$
answered Aug 7 at 19:48
mathcounterexamples.net
24.6k21653
Thanks! What about the condition of differentiability? $h$ is differentiability near $(1,2)$ because both $f$ and $vec g$ are differentiability? I mean, why can we apply the chain rule?
– manooooh
Aug 7 at 20:21
1
If $f$ is differentiable at $a$ and $g$ at $b=f(a)$ then $h =g circ f$ is differentiable at $a$ and its derivative is given by the chain rule. This is the chain rule theorem.
– mathcounterexamples.net
Aug 7 at 20:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Supposing that
$$D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright)$$
you also have $D_f(3,6)= (3,12)$.
Now you can write $(1.02,1.99)=(1+0.02,2.0-0.01)=(1,2)+(0.02,-0.01)=(1,2)+(h,k)$.
Based on that, you can use the chain rule (which you used in your question using coordinates), namely
$$D_h(1,2)=D_f(3,6) circ D_vec g(1,2)$$
To get
$$h(1.02,1.99)approx h(1,2)+D_h(1,2).(h,k)=45+(3,12)left[beginpmatrix2 &1\
3&5endpmatrixbeginpmatrix0.02\
-0.01endpmatrixright]=45.21$$
answered Aug 7 at 19:48
mathcounterexamples.net
24.6k21653
Thanks! What about the condition of differentiability? $h$ is differentiability near $(1,2)$ because both $f$ and $vec g$ are differentiability? I mean, why can we apply the chain rule?
– manooooh
Aug 7 at 20:21
1
If $f$ is differentiable at $a$ and $g$ at $b=f(a)$ then $h =g circ f$ is differentiable at $a$ and its derivative is given by the chain rule. This is the chain rule theorem.
– mathcounterexamples.net
Aug 7 at 20:34
add a comment |Â
up vote
2
down vote
accepted
Supposing that
$$D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright)$$
you also have $D_f(3,6)= (3,12)$.
Now you can write $(1.02,1.99)=(1+0.02,2.0-0.01)=(1,2)+(0.02,-0.01)=(1,2)+(h,k)$.
Based on that, you can use the chain rule (which you used in your question using coordinates), namely
$$D_h(1,2)=D_f(3,6) circ D_vec g(1,2)$$
To get
$$h(1.02,1.99)approx h(1,2)+D_h(1,2).(h,k)=45+(3,12)left[beginpmatrix2 &1\
3&5endpmatrixbeginpmatrix0.02\
-0.01endpmatrixright]=45.21$$
answered Aug 7 at 19:48
mathcounterexamples.net
24.6k21653
Thanks! What about the condition of differentiability? $h$ is differentiability near $(1,2)$ because both $f$ and $vec g$ are differentiability? I mean, why can we apply the chain rule?
– manooooh
Aug 7 at 20:21
1
If $f$ is differentiable at $a$ and $g$ at $b=f(a)$ then $h =g circ f$ is differentiable at $a$ and its derivative is given by the chain rule. This is the chain rule theorem.
– mathcounterexamples.net
Aug 7 at 20:34
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Supposing that
$$D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright)$$
you also have $D_f(3,6)= (3,12)$.
Now you can write $(1.02,1.99)=(1+0.02,2.0-0.01)=(1,2)+(0.02,-0.01)=(1,2)+(h,k)$.
Based on that, you can use the chain rule (which you used in your question using coordinates), namely
$$D_h(1,2)=D_f(3,6) circ D_vec g(1,2)$$
To get
$$h(1.02,1.99)approx h(1,2)+D_h(1,2).(h,k)=45+(3,12)left[beginpmatrix2 &1\
3&5endpmatrixbeginpmatrix0.02\
-0.01endpmatrixright]=45.21$$
answered Aug 7 at 19:48
mathcounterexamples.net
24.6k21653
Supposing that
$$D_vec g(1,2)=left(beginmatrix2&1\3&5endmatrixright)$$
you also have $D_f(3,6)= (3,12)$.
Now you can write $(1.02,1.99)=(1+0.02,2.0-0.01)=(1,2)+(0.02,-0.01)=(1,2)+(h,k)$.
Based on that, you can use the chain rule (which you used in your question using coordinates), namely
$$D_h(1,2)=D_f(3,6) circ D_vec g(1,2)$$
To get
$$h(1.02,1.99)approx h(1,2)+D_h(1,2).(h,k)=45+(3,12)left[beginpmatrix2 &1\
3&5endpmatrixbeginpmatrix0.02\
-0.01endpmatrixright]=45.21$$
answered Aug 7 at 19:48
mathcounterexamples.net
24.6k21653
answered Aug 7 at 19:48
mathcounterexamples.net
24.6k21653
answered Aug 7 at 19:48
mathcounterexamples.net
24.6k21653
answered Aug 7 at 19:48
mathcounterexamples.net
24.6k21653
24.6k21653
Thanks! What about the condition of differentiability? $h$ is differentiability near $(1,2)$ because both $f$ and $vec g$ are differentiability? I mean, why can we apply the chain rule?
– manooooh
Aug 7 at 20:21
1
If $f$ is differentiable at $a$ and $g$ at $b=f(a)$ then $h =g circ f$ is differentiable at $a$ and its derivative is given by the chain rule. This is the chain rule theorem.
– mathcounterexamples.net
Aug 7 at 20:34
add a comment |Â
Thanks! What about the condition of differentiability? $h$ is differentiability near $(1,2)$ because both $f$ and $vec g$ are differentiability? I mean, why can we apply the chain rule?
– manooooh
Aug 7 at 20:21
1
If $f$ is differentiable at $a$ and $g$ at $b=f(a)$ then $h =g circ f$ is differentiable at $a$ and its derivative is given by the chain rule. This is the chain rule theorem.
– mathcounterexamples.net
Aug 7 at 20:34
Thanks! What about the condition of differentiability? $h$ is differentiability near $(1,2)$ because both $f$ and $vec g$ are differentiability? I mean, why can we apply the chain rule?
– manooooh
Aug 7 at 20:21
Thanks! What about the condition of differentiability? $h$ is differentiability near $(1,2)$ because both $f$ and $vec g$ are differentiability? I mean, why can we apply the chain rule?
– manooooh
Aug 7 at 20:21
1
1
If $f$ is differentiable at $a$ and $g$ at $b=f(a)$ then $h =g circ f$ is differentiable at $a$ and its derivative is given by the chain rule. This is the chain rule theorem.
– mathcounterexamples.net
Aug 7 at 20:34
If $f$ is differentiable at $a$ and $g$ at $b=f(a)$ then $h =g circ f$ is differentiable at $a$ and its derivative is given by the chain rule. This is the chain rule theorem.
– mathcounterexamples.net
Aug 7 at 20:34
add a comment |Â
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1
You provide $quad D_vec g$. But in which $(x,y)$ value is it?
– mathcounterexamples.net
Aug 7 at 19:18
1
Using differentials means: using the approximation $f(x+delta)=f(x)+delta f'(x)$
– N74
Aug 7 at 19:20
@N74 thank you. Hmmm... and in the case for $2$ variables? Something like $h(x+delta_1,y+delta_2)$?
– manooooh
Aug 7 at 19:27
1
You have already the right expression for $h$
– N74
Aug 7 at 19:33
1
The vector form of my first comment... the derivative is replaced by the gradient in vector spaces
– N74
Aug 7 at 20:03